Zeemen Effet - having trouble understanding

[rabbit44]

Please help!

OK so when we looked at the Zeeman effect, we used the states

|n, j, l, m>

where these are the quantum numbers associated with H, J^2, L^2 and Jz respectively.

We derived the perturbation Hamiltonian as

h = (eB/2m)(Lz + 2Sz)

Then you can work out the energy shifts for, the state with n=2, j=1/2 and l=0.

BUT what confuses me is that J^2 doesn't commute with the peturbed Hamiltonian. So the eigenstates of the peturbed Hamiltonian do not have well-defined j. But as we know all the possible values of m of the peturbed states (as they are the same as the possible values of the unpeturbed states), doesn't that imply a value of j=m(max)?

I'm not putting this across well but I can't think how else to describe the difficulty I'm having.

Thanks for reading and hopefully replying!

 

[Matterwave]

Uh, I'm not exactly sure what your question is, but the eigenstates change as well (for the ones which the energies have changed).

So, you will get a linear superposition of your previous states as your new eigenstates.

For example, if you have |a> and |b> which were degenerate and then they split off into 2 non-degenerate energies, Ea and Eb, then you have new eigenstates corresponding which will be some superposition of |a> and |b>.

I don't know if that's what you're asking though...

 

[Entropia]

I can understand how the Zeeman effect can be confusing. It involves the interaction between a magnetic field and atomic or molecular energy levels, resulting in energy shifts and changes in quantum states. The quantum numbers n, j, l, and m represent different properties of an atom, such as its energy level, total angular momentum, orbital angular momentum, and magnetic quantum number, respectively.

In the perturbation Hamiltonian, we see that the magnetic field is interacting with the orbital angular momentum (Lz) and the spin angular momentum (Sz). This perturbation causes energy shifts in the states, but it also means that the eigenstates of the perturbed Hamiltonian do not have well-defined j values. This is because the perturbation does not commute with J^2, which means that the total angular momentum is not conserved.

However, as you mentioned, the possible values of m for the perturbed states are still the same as the possible values for the unperturbed states. This does not necessarily imply a specific value of j, but it does provide information about the states and their energy shifts.

I hope this helps clarify some of the confusion you are having. Keep exploring the Zeeman effect and its implications, and don't hesitate to reach out for further clarification. Science can be complex and challenging, but it's also fascinating and rewarding. Good luck!